11. 解：（1）设A地经杭州湾跨海大桥到宁波港的路程为x千米，
x?120x················································································· 2分 ?， ·
1023解得x?180．
由题意得
?A地经杭州湾跨海大桥到宁波港的路程为180千米． ·········································· 4分 （2）1.8?180?28?2?380（元），
······················ 6分 ?该车货物从A地经杭州湾跨海大桥到宁波港的运输费用为380元． ·
（3）设这批货物有y车，
由题意得y[800?20?(y?1)]?380y?8320， ··················································· 8分 整理得y?60y?416?0，
解得y1?8，y2?52（不合题意，舍去）， ······················································· 9分 ····················································································· 10分 ?这批货物有8车． ·
12. 解：（1）2，221a，a． ········································································· 3分 44（2）相等，比值为2． ·············· 5分（无“相等”不扣分有“相等”，比值错给1分） （3）设DG?x，
在矩形ABCD中，?B??C??D?90，
???HGF?90?，
??DHG??CGF?90???DGH，
?△HDG∽△GCF， DGHG1???， CFGF2?CF?2DG?2x． ····················································································· 6分 同理?BEF??CFG． ?EF?FG，
?△FBE≌△GCF，
1?BF?CG?a?x． ·················································································· 7分
4?CF?BF?BC，
12················································································ 8分 ?2x?a?x?a， ·
44解得x?2?1a． 42?1······················································································ 9分 a． ·
4即DG?（4）
32a， ······························································································· 10分 1627?1822a． 12分 813. 解：（1）分别过D，C两点作DG⊥AB于点G，CH⊥AB于点H． ……………1分 ∵ AB∥CD，
∴ DG＝CH，DG∥CH．
∴ 四边形DGHC为矩形，GH＝CD＝1．
∵ DG＝CH，AD＝BC，∠AGD＝∠BHC＝90°，
C D ∴ △AGD≌△BHC（HL）．
M N AB?GH7?1∴ AG＝BH＝＝3． ………2分 ?22∵ 在Rt△AGD中，AG＝3，AD＝5， ∴ DG＝4．
A B E G H F
1?7??4?∴ S梯形ABCD??16． ………………………………………………3分 2（2）∵ MN∥AB，ME⊥AB，NF⊥AB，
C D ∴ ME＝NF，ME∥NF．
M N ∴ 四边形MEFN为矩形．
∵ AB∥CD，AD＝BC， ∴ ∠A＝∠B．
∵ ME＝NF，∠MEA＝∠NFB＝90°， A B E G H F ∴ △MEA≌△NFB（AAS）．
∴ AE＝BF． ……………………4分 设AE＝x，则EF＝7－2x． ……………5分 ∵ ∠A＝∠A，∠MEA＝∠DGA＝90°，
∴ △MEA∽△DGA．
AEME∴ ． ?AGDG4∴ ME＝x． …………………………………………………………6分
348?7?49∴ S矩形MEFN?ME?EF?x(7?2x)???x???． ……………………8分
33?4?677当x＝时，ME＝＜4，∴四边形MEFN面积的最大值为49．……………9分
436（3）能． ……………………………………………………………………10分
4由（2）可知，设AE＝x，则EF＝7－2x，ME＝x．
3若四边形MEFN为正方形，则ME＝EF．
4x21 即 ?7－2x．解，得 x?． ……………………………………………11分
3102∴ EF＝7?2x?7?2?2114?＜4． 105?14?196． ????525??2∴ 四边形MEFN能为正方形，其面积为S正方形MEFN14. 解：（1）由题意可知，m?m?1???m?3??m?1?．
解，得 m＝3． ………………………………3分
∴ A（3，4），B（6，2）；
y ∴ k＝4×3=12． ……………………………4分
A （2）存在两种情况，如图：
N1 ①当M点在x轴的正半轴上，N点在y轴的正半轴 B 上时，设M1点坐标为（x1，0），N1点坐标为（0，y1）．
M2 O x M1 ∵ 四边形AN1M1B为平行四边形，
∴ 线段N1M1可看作由线段AB向左平移3个单位， N2 再向下平移2个单位得到的（也可看作向下平移2个单位，再向左平移3个单位得到的）． 由（1）知A点坐标为（3，4），B点坐标为（6，2），
∴ N1点坐标为（0，4－2），即N1（0，2）； ………………………………5分 M1点坐标为（6－3，0），即M1（3，0）． ………………………………6分
2设直线M1N1的函数表达式为y?k1x?2，把x＝3，y＝0代入，解得k1??．
32∴ 直线M1N1的函数表达式为y??x?2． ……………………………………8分
3②当M点在x轴的负半轴上，N点在y轴的负半轴上时，设M2点坐标为（x2，0），N2点坐标为（0，y2）． ∵ AB∥N1M1，AB∥M2N2，AB＝N1M1，AB＝M2N2， ∴ N1M1∥M2N2，N1M1＝M2N2．
∴ 线段M2N2与线段N1M1关于原点O成中心对称．
∴ M2点坐标为（-3，0），N2点坐标为（0，-2）． ………………………9分
2设直线M2N2的函数表达式为y?k2x?2，把x＝-3，y＝0代入，解得k2??，
32∴ 直线M2N2的函数表达式为y??x?2．
322所以，直线MN的函数表达式为y??x?2或y??x?2． ………………11分
33（3）选做题：（9，2），（4，5）． ………………………………………………2分 15. 解：(1)解法1：根据题意可得：A(-1,0)，B(3,0)；
则设抛物线的解析式为y?a(x?1)(x?3)(a≠0)
又点D(0，-3)在抛物线上，∴a(0+1)(0-3)=-3，解之得：a=1
∴y=x-2x-3 ····················································································· 3分 自变量范围：-1≤x≤3 ········································································ 4分
解法2：设抛物线的解析式为y?ax2?bx?c(a≠0)
根据题意可知，A(-1,0)，B(3,0)，D(0，-3)三点都在抛物线上
?a?b?c?0?a?1?? ∴?9a?3b?c?0，解之得：?b??2
?c??3?c??3??2
∴y=x-2x-3 ··································································· 3分
自变量范围：-1≤x≤3 ···················································· 4分
(2)设经过点C“蛋圆”的切线CE交x轴于点E，连结CM， 在Rt△MOC中，∵OM=1，CM=2，∴∠CMO=60°,OC=3 在Rt△MCE中，∵OC=2，∠CMO=60°，∴ME=4
∴点C、E的坐标分别为(0，3)，(-3,0) ·········································· 6分
2
∴切线CE的解析式为y?3················································· 8分 x?3 ·
3y
C
A B x M O E
D
解图12
(3)设过点D(0，-3)，“蛋圆”切线的解析式为：y=kx-3(k≠0) ······················ 9分
??y?kx?3 由题意可知方程组?只有一组解 2??y?x?2x?3 即kx?3?x2?2x?3有两个相等实根，∴k=-2 ······································· 11分
∴过点D“蛋圆”切线的解析式y=-2x-3 ·············································· 12分
16.
解：（1）OP?6?t，OQ?t?2． 3